Eliminate θ from the equation xsinθ−ycosθ=x2+y2 and cos2θa2+sin2θb2=1x2+y2

xsinθ−ycosθ=x2+y2→(1) And cos2θa2+sin2θb2=1x2+y2→(2) Squaring on both sides of (i) x2sin2θ+y2cos2θ−2xysinθ.cosθ=x2+y2 x2(1−cos2θ)+y2(1−sin2θ)−2xysinθ.cosθ=x2+y2 x2−x2cos2θ+y2−y2sin2θ−2xysinθ.cosθ=x2+y2 x2+y2−(xcosθ+ysinθ)2=x2+y2 (xcosθ+ysinθ)2=0 xcosθ=−ysinθ tanθ=−xy tanθ is negative in Q2 & Q4 Case(1): θ∈Q2 ∴cosθ=−yx2+y2 and sinθ=xx2+y2 Substituting these values of cosθ and sinθ in (ii) 1a2×y2x2+y2+1b2×x2x2+y2=1x2+y2 ⇒x2a2+y2a2=1 Case (2): θ∈Q4 ∴cosθ=−yx2+y2 and sinθ=−xx2+y2 Substituting these values of cosθ and sinθ in (ii) 1a2×y2x2+y2+1b2×x2x2+y2=1x2+y2 ⇒x2b2+y2a2=1

Eliminate θ from the equation xsinθ−ycosθ=x2+y2 and cos2θa2+sin2θb2=1x2+y2

xsinθ−ycosθ=x2+y2→(1) And cos2θa2+sin2θb2=1x2+y2→(2) Squaring on both sides of (i) x2sin2θ+y2cos2θ−2xysinθ.cosθ=x2+y2 x2(1−cos2θ)+y2(1−sin2θ)−2xysinθ.cosθ=x2+y2 x2−x2cos2θ+y2−y2sin2θ−2xysinθ.cosθ=x2+y2 x2+y2−(xcosθ+ysinθ)2=x2+y2 (xcosθ+ysinθ)2=0 xcosθ=−ysinθ tanθ=−xy tanθ is negative in Q2 & Q4 Case(1): θ∈Q2 ∴cosθ=−yx2+y2 and sinθ=xx2+y2 Substituting these values of cosθ and sinθ in (ii) 1a2×y2x2+y2+1b2×x2x2+y2=1x2+y2 ⇒x2a2+y2a2=1 Case (2): θ∈Q4 ∴cosθ=−yx2+y2 and sinθ=−xx2+y2 Substituting these values of cosθ and sinθ in (ii) 1a2×y2x2+y2+1b2×x2x2+y2=1x2+y2 ⇒x2b2+y2a2=1

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Eliminate $θ$ from the equation $x \sin θ - y \cos θ = \sqrt{x^{2} + y^{2}}$ and $\frac{\cos^{2} θ}{a^{2}} + \frac{\sin^{2} θ}{b^{2}} = \frac{1}{x^{2} + y^{2}}$

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\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2}} = 1

\frac{x}{a} + \frac{y}{b} = 1

\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1

answer is D.

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Detailed Solution

x \sin θ - y \cos θ = \sqrt{x^{2} + y^{2}} \to (1)

And

$\frac{\cos^{2} θ}{a^{2}} + \frac{\sin^{2} θ}{b^{2}} = \frac{1}{x^{2} + y^{2}} \to (2)$

Squaring on both sides of (i)

$x^{2} \sin^{2} θ + y^{2} \cos^{2} θ - 2 x y \sin θ . \cos θ = x^{2} + y^{2}$

$x^{2} (1 - \cos^{2} θ) + y^{2} (1 - \sin^{2} θ) - 2 x y \sin θ . \cos θ = x^{2} + y^{2}$

$x^{2} - x^{2} \cos^{2} θ + y^{2} - y^{2} \sin^{2} θ - 2 x y \sin θ . \cos θ = x^{2} + y^{2}$

$x^{2} + y^{2} - {(x \cos θ + y \sin θ)}^{2} = x^{2} + y^{2}$

${(x \cos θ + y \sin θ)}^{2} = 0$

$x \cos θ = - y \sin θ$

$\tan θ = \frac{- x}{y}$

Question Image

$\tan θ$ is negative in $Q_{2} & Q_{4}$

Case(1): $θ \in Q_{2}$

$∴ \cos θ = \frac{- y}{\sqrt{x^{2} + y^{2}}}$ and $\sin θ = \frac{x}{\sqrt{x^{2} + y^{2}}}$

Substituting these values of $\cos θ$ and $\sin θ$ in (ii)

$\frac{1}{a^{2}} \times \frac{y^{2}}{x^{2} + y^{2}} + \frac{1}{b^{2}} \times \frac{x^{2}}{x^{2} + y^{2}} = \frac{1}{x^{2} + y^{2}}$ $\Rightarrow \frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2}} = 1$

Case (2): $θ \in Q_{4}$

$∴ \cos θ = \frac{- y}{\sqrt{x^{2} + y^{2}}}$ and $\sin θ = \frac{- x}{\sqrt{x^{2} + y^{2}}}$

Substituting these values of $\cos θ$ and $\sin θ$ in (ii)

$\frac{1}{a^{2}} \times \frac{y^{2}}{x^{2} + y^{2}} + \frac{1}{b^{2}} \times \frac{x^{2}}{x^{2} + y^{2}} = \frac{1}{x^{2} + y^{2}}$ $\Rightarrow \frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1$